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FREQUENCY of CYCLIC CODES AN EXAMPLE of a CYCLIC CODE Comparing with linear codes, cyclic codes are quite scarce. For example, there are 11 811 linear [7,3] binary codes, but only two of them are cyclic. Trivial cyclic codes. For any field F and any integer n ≥ 3 there are always the following cyclic codes of length n over F : No-information code - code consisting of just one all-zero codeword. Repetition code - code consisting of all codewords (a, a, . . . ,a) for a ∈ F . Single-parity-check code - code consisting of all codewords with parity 0. No-parity code - code consisting of all codewords of length n For some cases, for example for n = 19 and F = GF (2), the above four trivial cyclic codes are the only cyclic codes. Is the code with the following generator matrix cyclic? 2 1 0 1 1 1 0 3 0 G = 40 1 0 1 1 1 05 0 0 1 0 1 1 1 It is. It has, in addition to the codeword 0000000, the following codewords c 1 = 1011100 c 1 + c 2 = 1110010 c 2 = 0101110 c 1 + c 3 = 1001011 c 1 + c 2 + c 3 = 1100101 and it is cyclic because the right shifts have the following impacts c 1 → c 2 , c 1 + c 2 → c 2 + c 3 , c 2 → c 3 , c 3 = 0010111 c 2 + c 3 = 0111001 c 3 → c 1 + c 3 c 1 + c 3 → c 1 + c 2 + c 3 , c 2 + c 3 → c 1 c 1 + c 2 + c 3 → c 1 + c 2 prof. Jozef Gruska IV054 3. Cyclic codes 5/71 POLYNOMIALS over GF(q) prof. Jozef Gruska IV054 3. Cyclic codes 6/71 NOTICE A codeword of a cyclic code is usually denoted a 0 a 1 . . . a n−1 and to each such a codeword the polynomial will be associated – am ingenious idea!!. a 0 + a 1 x + a 2 x 2 + . . . + a n−1 x n−1 NOTATION: F q [x] will denote the set of all polynomials f (x) over GF (q). deg(f (x)) = the largest m such that x m has a non-zero coefficient in f (x). Multiplication of polynomials If f (x), g(x) ∈ FQ[x], then deg(f (x)g(x)) = deg(f (x)) + deg(g(x)). Division of polynomials For every pair of polynomials a(x), b(x) ≠ 0 in F q [x] there exists a unique pair of polynomials q(x), r(x) in F q [x] such that a(x) = q(x)b(x) + r(x), deg(r(x)) < deg(b(x)). Example Divide x 3 + x + 1 by x 2 + x + 1 in F 2 [x]. Definition Let f (x) be a fixed polynomial in F q [x]. Two polynomials g(x), h(x) are said to be congruent modulo f (x), notation if g(x) − h(x) is divisible by f (x). g(x) ≡ h(x) (mod f (x)), prof. Jozef Gruska IV054 3. Cyclic codes 7/71 A code C of the words of length n is a set of codewords of length n a 0 a 1 a 2 . . . a n−1 or C can be seen as a set of polynomials of the degree (at most) n − 1 a 0 + a 1 x + a 2 x 2 + . . . + a n−1 x n−1 prof. Jozef Gruska IV054 3. Cyclic codes 8/71
RINGS of POLYNOMIALS For any polynomial f (x), the set of all polynomials in F q [x] of degree less than deg(f (x)), with addition and multiplication modulo f (x), forms a ring denoted F q [x]/f (x). Example Calculate (x + 1) 2 in F 2 [x]/(x 2 + x + 1). It holds (x + 1) 2 = x 2 + 2x + 1 ≡ x 2 + 1 ≡ x (mod x 2 + x + 1). How many elements has F q [x]/f (x)? Result |F q [x]/f (x)| = q deg(f (x)) . Example Addition and multiplication tables for F 2 [x]/(x 2 + x + 1) + 0 1 x 1+x 0 0 1 x 1+x 1 1 0 1+x x x x 1+x 0 1 1+x 1+x x 1 0 • 0 1 x 1+x 0 0 0 0 0 1 0 1 x 1+x x 0 x 1+x 1 1+x 0 1+x 1 x Definition A polynomial f (x) in F q [x] is said to be reducible if f (x) = a(x)b(x), where a(x), b(x) ∈ F q [x] and deg(a(x)) < deg(f (x)), deg(b(x)) < deg(f (x)). If f (x) is not reducible, then it is said to be irreducible in F q [x]. Theorem The ring F q [x]/f (x) is a field if f (x) is irreducible in F q [x]. FIELD R n , R n = F q [x]/(x n − 1) Computation modulo x n − 1 in the field R n = F q [x]/(x n − 1) Since x n ≡ 1 (mod (x n − 1)) we can compute f (x) mod (x n − 1) by replacing, in f (x), x n by1, x n+1 by x, x n+2 by x 2 , x n+3 by x 3 , . . . Replacement of a word by a polynomial is of large importance because w = a 0 a 1 . . . a n−1 p(w) = a 0 + a 1 x + . . . + a n−1 x n−1 multiplication of p(w) by x in R n corresponds to a single cyclic shift of w x(a 0 + a 1 x + . . . a n−1 x n−1 ) = a n−1 + a 0 x + a 1 x 2 + . . . + a n−2 x n−1 prof. Jozef Gruska IV054 3. Cyclic codes 9/71 An ALGEBRAIC CHARACTERIZATION of CYCLIC CODES Theorem A binary code C of words of length n is cyclic if and only if it satisfies two conditions (i) a(x), b(x) ∈ C ⇒ a(x) + b(x) ∈ C (ii) a(x) ∈ C, r(x) ∈ R n ⇒ r(x)a(x) ∈ C Proof (1) Let C be a cyclic code. C is linear ⇒ (i) holds. (ii) If a(x) ∈ C, r(x) = r 0 + r 1 x + . . . + r n−1 x n−1 then r(x)a(x) = r 0 a(x) + r 1 xa(x) + . . . + r n−1 x n−1 a(x) is in C by (i) because summands are cyclic shifts of a(x). (2) Let (i) and (ii) hold Taking r(x) to be a scalar the conditions (i) and (ii) imply linearity of C. Taking r(x) = x the conditions (i) and (ii) imply cyclicity of C. prof. Jozef Gruska IV054 3. Cyclic codes 10/71 CONSTRUCTION of CYCLIC CODES Notation For any f (x) ∈ R n , we can define 〈f (x)〉 = {r(x)f (x) | r(x) ∈ R n } (with multiplication modulo x n − 1) to be a set of polynomials - a code. Theorem For any f (x) ∈ R n , the set 〈f (x)〉 is a cyclic code (generated by f ). Proof We check conditions (i) and (ii) of the previous theorem. (i) If a(x)f (x) ∈ 〈f (x)〉 and also b(x)f (x) ∈ 〈f (x)〉, then (ii) If a(x)f (x) ∈ 〈f (x)〉, r(x) ∈ R n , then a(x)f (x) + b(x)f (x) = (a(x) + b(x))f (x) ∈ 〈f (x)〉 r(x)(a(x)f (x)) = (r(x)a(x))f (x) ∈ 〈f (x)〉 Example let C = 〈1 + x 2 〉, n = 3, q = 2. In order to determine C we have to compute r(x)(1 + x 2 ) for all r(x) ∈ R 3 . Result R 3 = {0, 1, x, 1 + x, x 2 , 1 + x 2 , x + x 2 , 1 + x + x 2 }. C = {0, 1 + x, 1 + x 2 , x + x 2 } C = {000, 110, 101, 011} prof. Jozef Gruska IV054 3. Cyclic codes 11/71 prof. Jozef Gruska IV054 3. Cyclic codes 12/71
Page 1: CHAPTER 3: CYCLIC CODES, CHANNEL CO
Page 5 and 6: EXAMPLE The task is to determine al
Page 7 and 8: MULTIPLICATION of POLYNOMIALS by SH
Page 9 and 10: POLYNOMIAL CODES BCH CODES and REED
Page 11 and 12: CHANNEL (STREAM) CODING II CONVOLUT
Page 13 and 14: CONCATENATED CODES - I CONCATENATED
Page 15 and 16: DECODING and PERFORMANCE of TURBO C
Page 17 and 18: APPENDIX APPLICATIONS of REED-SOLOM

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